<p>Page 473 #8
(Diagram of triangle needed)
In the figure above, the is the value of c in terms of a and b</p>
<p>Page 475 #16:
Let the function h be defined by h(x) = 14 + x^2/4
If h(2m) = 9m, what is one possible value of m?</p>
<p>Page 476 #17
(table needed)
A merchant sells 3 types of clocks that chime as indicated by the check marks in the table above. what is the total number of chimes of the inventory of clocks in the 90 minute period from 7:15 to 8:45</p>
<p>Page 476 #18
If 5 cards A B C D E are placed in a row so that C is never at either end, how many different arrangements are possible?</p>
<p>Consider the 3 triangles as a trapezoid. The sum of the interior angles is 360.</p>
<p>b + c + a + b + b + a = 360.
c + 3b + 2a = 360
c = 360 - 2a - 3b</p>
<h1>16</h1>
<p>Plug in h(2m). You get m^2 + 14 = 9m. So, m^2 - 9m + 14 = 0. Solve it in a graphing calculator for m = 2 or 7</p>
<h1>17</h1>
<p>Look at each type.</p>
<p>Type A = Chime 8 times on the 8th hour + once every half hour.
Type A = (eight chimes at 8 oclock) + 2 chimes (7:30, 8:30)</p>
<p>Each clock chimes 10 distinct times, so 100 chimes. </p>
<p>Type B = (eight chimes at 8 oclock).</p>
<p>5 * 8 = 40 chimes</p>
<p>Type C = (one chime at 8) + (two chimes at 7:30, 8:30). </p>
<p>3 * 3 = 9</p>
<p>149 total. </p>
<h1>18</h1>
<p>There are 5! ways to put 5 cards in any order, or 120 ways. We want to eliminate all possibilities where the gray is at the ends. If the list starts as gray, there are 4! ways to put the other cards, or 24 ways. If gray is at the end, there are also 4! ways to put the other cards. </p>
<p>Last one: To get the 5! you need to understand that any 5 cards could go in the first “slot,” but then only one of 4 cards can go in the second slot, because the first card is already put on the table. Then only 3 cards are left to be chosen from for the next slot, and so forth. Multiply these together to get the total combos possible. You get 5x4x3x2x1 or 5!. </p>
<p>Let’s suppose C is the first card on the table. Again, any one of the 4 cards can be put down next, then any of the 3 remaining cards, etc. So there are 4! possible combos with C leading. Same thing if C is the last card on the table - 4!.</p>
<p>The question asked the possible combos if C was not at either end. So, just subtract. 5! - (4!+4!) or 120-48 = 72.</p>
<p>Anamoly, can you explain the rationale for choosing “(eight chimes at 8 oclock) + 2 chimes (7:30, 8:30)” I do not understand that part. If you may explain it, then it will be appreciated Thank you</p>